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goldenfox [79]
2 years ago
14

A sound wave, generated at a frequency of 440 hertz has a wavelength of 2.3 meters as it travels through a solid material. The a

pproximate speed of the wave is ____ m/sec.
Physics
1 answer:
Jlenok [28]2 years ago
8 0

The approximate speed of the sound wave traveling through the solid material is 1012m/s.

<h3>Wavelength, Frequency and Speed</h3>

Wavelength is simply the distance over which the shapes of waves are repeated. It is the spatial period of a periodic wave.

From the wavelength, frequency and speed relation,

λ = v ÷ f

Where λ is wavelength, v is velocity/speed and f is frequency.

Given the data in the question;

  • Frequency of sound wave f = 440Hz = 440s⁻¹
  • Wavelength of the wave λ = 2.3m
  • Speed of the wave v = ?

To determine the approximate speed of the wave, we substitute our given values into the expression above.

λ = v ÷ f

2.3m = v ÷ 440s⁻¹

v = 2.3m × 440s⁻¹

v = 1012ms⁻¹

v = 1012m/s

Therefore, the approximate speed of the sound wave traveling through the solid material is 1012m/s.

Learn more about Speed, Frequency and Wavelength here: brainly.com/question/27120701

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Feliz [49]

Answer:

The final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

Explanation:

Taking north direction as positive y axis and east direction as positive x axis .

Given:

Mass of first car is, m_1=1000\ kg

Initial velocity of first car is, u_1=30\vec{j}\ m/s

Mass of second car is, m_2=1250\ kg

Initial velocity of second car is,

Let the combined final velocity after collision be 'v' m/s with as components of final velocity along east and north directions respectively.

Now, as the net external force is zero, momentum is conserved for the two car system along the east and north directions.

Conserving momentum along the east direction, we have:

Initial momentum = Final momentum

m_1u_{1x}+m_2u_{2x}=(m_1+m_2)v_x\\\\0+1250\times 39=(1000+1250)v_x\\\\v_x=\frac{48750}{2250}\\\\v_x=\frac{65}{3} m/s

There is no component of initial velocity for first car in east direction, as it is moving in the north direction. So,

Now, conserving momentum along the north direction, we have:

Initial momentum = Final momentum

m_1u_{1y}+m_2u_{2y}=(m_1+m_2)v_y\\\\1000\times 30+0=(1000+1250)v_y\\\\v_y=\frac{30000}{2250}\\\\v_y=\frac{40}{3}\ m/s

There is no component of initial velocity for second car in north direction, as it is moving in the east direction. So, u_{2y}=0.

The magnitude of final velocity is given as:

|\vec{v}|=\sqrt{(v_x)^2+(v_y)^2}\\\\|\vec{v}|=\sqrt{(\frac{65}{3})^2+(\frac{40}{3})^2}\\\\|\vec{v}|=\sqrt{\frac{5825}{9}}=25.44\ m/s

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{\frac{40}{3}}{\frac{65}{3}})=31.61^\circ

So, the final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

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